******************************************************************** Section 4.7, pp. 123 - 129: Types of intervals of convergence: Interval of Series Convergence \sum n^n x^n {0} \sum x^n / n! R \sum x^n / n [-1,1) \sum (-1)^{n+1} x^n / n (-1,1] \sum x^n / n^2 [-1,1] We can also make a power series whose interval of convergence is (-1,1), but I don't know such an easy way to do it. Here's one example: Let e_1 := 1, e_2 := 1, e_3 := 1, e_4 := -1, and the recursively define e_{n+4} := e_n. Thus the sequence e_i is 1,1,1,-1,1,1,1,-1,1,1,1,-1,1,1,1,-1,1,1,1,-1,1,1,1,-1,... We have 1 + (1/3) + (1/5) + (1/7) + (1/9) + (1/11) + ... = \infty, because it dominates term-by-term one half of the harmonic series, and (1/2) - (1/4) + (1/6) - (1/8) + (1/10) - (1/12) + ... converges because of the Alternating-Series Test, Theorem 6.6(i), pp. 118-119. If we add these two sequences, then we find that \sum e_n / n = \infty. If we subtract the first from the second, then we find that \sum (-1)^n e_n / n = -\infty. Thus we see that \sum e_n x^n / n coverges neither at x=1 nor at x=-1. The interval of covergence of \sum e_n x^n / n is (-1,1). ******************************************************************** On p. 128, problem 7.8(h), it would be clearer to replace 2/n by 2 - n. ******************************************************************** On p. 122, problem 6.15, there are two parts labeled "b" and none labeled "c". ******************************************************************** Fei Xu's (the grader's) solution to Exercise 6.26, p. 74: For all n, we have n+3 n (n+3)a_{n-1} n a_n = --- a_{n-1} + ------------- \ge 2 \sqrt{ ------------ + ------------ } 2n (n+1) a_{n-1} 2n (n+1)a_{n-1} n+3 = 2 \sqrt{ ------ } > 2. In the \ge step, we used 2(n+1) the arithmetic-geometric mean inequality. n+3 \sqrt{2} Define b_n = --- b_{n-1} + --------, with b_0=2. 2n 2 By arithmetic-geometric mean inequality, we get b_n \ge \sqrt{2}/2. We have n+3 \sqrt{2} n b_n - a_n = --- (b_{n-1} - a_{n-1}) + -------- - ------------ 2n 2 (n+1)a_{n-1}. \sqrt{2} n Since -------- - ------------ is positive, so induction shows that: 2 (n+1)a_{n-1} for all n, we have b_n > a_n. n+3 n+2 n+3 n+2 We have b_n - b_{n-1} = --- b_{n-1} - ------ b_{n-2} and --- < ------ 2n 2(n-1) 2n 2(n-1). We calculate that 35 123 13 b_4 = -- + --- \sqrt{2} < 5 + -- \sqrt{2} = b_3. 8 64 8 Therefore, induction shows, for all n \ge 3, that b_{n+1} < b_n. Then, as b_n is bounded below, we see that b_n is convergent. Let x := \lim b_n. Then, from the definition of b_n, we see that x = (1/2) x + ((\sqrt{2}) / 2). Solving, we find that x = \sqrt{2}. So b_n --> \sqrt{2}. Recall that, for all n, we have \sqrt{2} < a_n < b_n. Then a_n --> \sqrt{2}. ******************************************************************** The geometric mean of the ratios test. Lemma. Let \sum a_n be a series with positive terms. Let r_n := a_{n+1} / a_n. Let s_n := (r_1 ... r_n)^{1/n} be the n'th root of the product of r_1, ..., r_n. Let L < 1. Assume for all sufficiently large n, that s_n \le L. Then \sum a_n converges. Proof: We have [a_1][(s_n)^n] = a_n, so (a_n)^{1/n} = [(a_1)^{1/n}][s_n]. Let L' := (1+L)/2. Since (a_1)^{1/n} --> 1 and since, for all n, we have s_n < L, we conclude, for all sufficiently large n, that (a_n)^{1/n} < L'. The ratio test then tells us that \sum a_n converges. QED In fact, the series for which the geometric mean of the ratios test will work are exactly the same as those for which the root test works. ******************************************************************** The arithmetic mean of the ratios test. Lemma. Let \sum a_n be a series with positive terms. Let r_n := a_{n+1} / a_n. Let t_n := (1/n)(r_1 + ... + r_n) be the average of r_1,...,r_n. Assume, for all sufficiently large n, that t_n \le L. Then \sum a_n converges. Proof: For all n, let x_n := \ln r_n. For all n, let s_n := (r_1 ... r_n)^{1/n}. Because \ln is concave down, it follows, for all n, that \ln t_n \ge (1/n)(x_n + ... + x_n). Then, for all n, \ln t_n \ge \ln s_n. Then, for all n, t_n \ge s_n. then, for all sufficiently large n, we have s_n \le L. So, by the geometric mean of the ratios test (above), we are done. QED ******************************************************************** The root test is better than the arithmetic mean of the ratios test: Let z := e^{-1/2} = 0.606530659... x := 10(1-z) = 3.934693403... Then z < 1 < x. We compute that z + (x/50) = .685224527... < 3/4. Let a_n be the sequence which is all zeroes except: if n = 10, then a_n = x, if n = 110, then a_n = 10x, if n = 1110, then a_n = 100x, if n = 11110, then a_n = 1000x, etc. Let b_n be the constant sequence x,x,x,x,... For all n, let c_n = a_n + b_n. The basic feature of c_n is that its averages do not converge, but the averages of \ln c_n do converge, to -1/2. Below, we will show that the arithmetic mean of the ratios test does not work on the series c_1 + c_1 c_2 + c_1 c_2 c_3 + ... but that the root test does show that that this series converges. Proof that the arithmetic mean of the ratios test does not work: > If n \in {10, 110, 1110, 11110, ...}, then > > c_1 + ... + c_n > --------------- = 1. > n > > If n = 10 - 1 = 9, then > > c_1 + ... + c_n > --------------- = z < z + (x/50) < 3/4. > n > > If n = 110 - 1 = 109, then > > c_1 + ... + c_n x > --------------- = z + - < z + (x/50) < 3/4. > n n > > If n = 1110 - 1 = 1109, then > > c_1 + ... + c_n 11x 20 x > --------------- = z + --- < z + ---- = z + (x/50) < 3/4. > n n 1000 > > If n = 11110 - 1 = 11109, then > > c_1 + ... + c_n 111 x 200 x > --------------- = z + ----- < z + ----- = z + (x/50) < 3/4. > n n 10000 > > Etc. > > In general, if n+1 \in {10, 110, 1110, 11110, ...}, then > > c_1 + ... + c_n > --------------- < 3/4. > n > > In particular, we see that > > c_1 + ... + c_n > --------------- > n > > does not have a limit, because it moves back and forth between 1 > and numbers less than 3/4. > > It follows from this that > > c_2 + ... + c_{n+1} > ------------------ > n > > does not have a limit, which implies that the arithmetic mean of the > ratios test does not work on the series > c_1 + c_1 c_2 + c_1 c_2 c_3 + ... Proof that the root test does work: > Let \alpha_n be the sequence which is all zeroes except: > if n = 10, then \alpha_n = (1/2) + \ln x + \ln 10, > if n = 110, then \alpha_n = (1/2) + \ln x + 2 \ln 10, > if n = 1110, then \alpha_n = (1/2) + \ln x + 3 \ln 10, > if n = 11110, then \alpha_n = (1/2) + \ln x + 4 \ln 10, > etc. > > For all n, we have \beta_n = \ln c_n. > For all n, we have \ln c_n \le \gamma_n. For example, if n = 1110, then > \ln c_n = \ln(a_n + b_n) = \ln(100 x + z) \le \ln(101 x) = > = \ln 101 + \ln x \le \ln x + \ln 1000 = \ln x + 3 \ln 10 = \gamma_n. > So, for all n, we have \beta_n \le \ln c_n \le \gamma_n. > > Because of the lacunary nature of \alpha_n, it is easy to show that > > \alpha_1 + ... + \alpha_n > ------------------------- ---> 0. > n > > Let \beta_n be the constant sequence -1/2, -1/2, -1/2, .... Then > > \beta_1 + ... + \beta_n > ----------------------- ---> -1/2. > n > > For all n, let \gamma_n := \alpha_n + \beta_n. Then > > \gamma_1 + ... + \gamma_n > ------------------------- ---> -1/2. > n > > So, since, for all n, we have \beta_n \le \ln c_n \le \gamma_n, > we conclude that > > (\ln c_1) + ... + (\ln c_n) > --------------------------- ---> -1/2. > n > > Exponentiating, we get (c_1 ... c_n)^{1/n} ---> z < 1. > Therefore the root test shows that the series > c_1 + c_1 c_2 + c_1 c_2 c_3 + ... > does converge. ******************************************************************** Typographical error on p. 106, line 16. n We want a_n := ----------------------, \sqrt{ n^4 + n^3 + 17} n not a_n := \sum ----------------------. \sqrt{ n^4 + n^3 + 17} ******************************************************************** A good lemma for proving Theorem 6.6, p. 118: Lemma. If a_1 \ge a_2 \ge ... \ge a_n \ge 0 and if S = \sum (-1)^{n+1} a_n = a_1 - a_2 + a_3 - a_4 + ... \pm a_n, then 0 \le S \le a_1. Using this we can show that if \sum b_n is an alternating series whose terms tend to zero and whose terms have nonincreasing absolute value, then the partial sums of \sum b_n are Cauchy, so \sum b_n is convergent. ******************************************************************** In the version of Theorem 4.6, p. 104 presented in class we had: Thm. Suppose a_n and b_n are both sequences of positive real numbers. Assume that a_n/b_n is convergent and that \sum b_n is convergent. Then \sum a_n is convergent. It is important in this result that a_n and b_n be sequences of positive real numbers; it is not true for arbitrary sequences of real numbers: For example, for n\ge 1, let a_n := 1/n and b_n := ((-1)^n)/(\sqrt{n}). Then a_n/b_n = b_n --> 0. By (i) of Theorem 6.6, p. 118, we see that \sum b_n is convergent. However, \sum a_n is the harmonic series and is therefore not convergent; see Theorem 3.4, p. 98. ******************************************************************** Argument that the root test is "better" than the ratio test: ******** Example where the root test works, but the ratio test doesn't: 1/4 + 1/4 + 1/16 + 1/16 + 1/64 + 1/64 + ... Note that the ratios are 1 , 1/4 , 1 , 1/4 , 1 , ... So the ratios don't have a limit, and so the ratio test gives no answer. For even n, the nth root of the nth term is exactly 1/2. For odd n, one can argue that the nth root of the nth term approaches 1/2. Thus the root test works and shows that the series converges. ********* Lemma that shows that if the ratio test works, then the root test works as well: Lemma. Let a_n be a sequence of positive numbers, defined for n \ge 0. Then a_{n+1}/a_n --> L implies (a_n)^{1/n} --> L. Pf: Letting b_n := \ln a_n and K := \ln L, we want to show that b_{n+1} - b_n --> K implies (1/n)b_n --> K. Changing n to n-1 does not change the limit, so b_{n+1} - b_n --> K is equivalent to b_n - b_{n-1} --> K. We therefore wish to show that b_n - b_{n-1} --> K implies (1/n)b_n --> K. Letting c_n := b_n - b_{n-1}, for n \ge 1, we have c_1 + ... + c_n = b_n - b_0. We therefore want to show that c_n --> K implies (1/n)(c_1 + ... + c_n + b_0) --> K. Since (1/n) b_0 --> 0, we want to show that c_n --> K implies (1/n)(c_1 + ... + c_n) --> K. Let d_n := c_n - K. Then (1/n)(d_1 + ... + d_n) = (1/n)(c_1 + ... + c_n) - K. Thus, we wish to show that d_n --> 0 implies (1/n)(d_1 + ... + d_n) --> 0. Assume that d_n --> 0. Given \epsilon > 0. We wish to show that there is n_0 such that, for all n \ge n_0, | (1/n)(d_1 + ... + d_n) | < \epsilon. Choose k such that, for all n\ge k, | d_n | < \epsilon / 2. Let S := | d_1 + ... + d_k |. Choose m such that, for all n\ge m, S / n < \epsilon / 2. Let n_0 := max{k+1,m}. Given n \ge n_0. We wish to show that | (1/n)(d_1 + ... + d_n) | < \epsilon. By the triangle inequality, we have | (1/n)(d_1 + ... + d_n) | \le (1/n)( S + |d_{k+1}| + ... + |d_n| ) Since n \ge n_0 \ge m, we conclude that S / n < \epsilon / 2, By the choice of k, we have |d_{k+1}| < \epsilon / 2 , ... , |d_n| < \epsilon / 2. Then |d_{k+1}| + ... + |d_n| < (n-k) \epsilon / 2 < n \epsilon / 2. Then | (1/n)(d_{k+1} + ... + d_n) | \le (1/n)( |d_{k+1}| + ... + |d_n| ) < \epsilon / 2. Then | (1/n)(d_1 + ... + d_n) | \le (1/n)( S + |d_{k+1}| + ... + |d_n| ) \le (S / n) + (1/n)( |d_{k+1}| + ... + |d_n| ) < (\epsilon / 2) + (\epsilon / 2) = \epsilon. QED ******************************************************************** Exercise 1.12(e) p. 86. \int x^2/(x^4+1) dx. Let a := \sqrt{2}. We get x^4 + 1 = (x^2 + ax + 1)(x^2 - ax + 1) Division algorthm shows: eqn 1: [ x^2 + ax + 1 ] - [ x^2 - ax + 1 ] = 2ax and then eqn 2: [ x^2 - ax + 1 ] - [2ax][x/(2a) - (1/2)] = 1. Plugging in the LHS of the eqn 1 into the [2ax] of eqn 2 and factoring out x^2 + ax + 1 and x^2 - ax + 1 from what appears, we get [x^2 - ax + 1][x/(2a) + (1/2)] + [x^2 + ax + 1][-x/(2a) + (1/2)] = 1. Dividing by (x^2 + ax + 1)(x^2 - ax + 1), we get (1/2) + x/(2a) (1/2) - x/(2a) 1 ---------------- + ----------------- = ------- x^2 + ax + 1 x^2 - ax + 1 x^4 + 1. If we multiply through by x^2 and then apply the division algorithm to the two terms in the LHS of the resulting equation in order to make sure that every quotient of polynomials has the property that its numerator has smaller degree than its denominator, then we get -x/(2a) x/(2a) x^2 eqn 3: ------------ + ------------ = ------- x^2 + ax + 1 x^2 - ax + 1 x^4 + 1. If we can integrate all four of the following rational functions 2x + a 2x - a 1 1 ------------ and ------------ and ------------ and ------------ x^2 + ax + 1 x^2 - ax + 1 x^2 + ax + 1 x^2 - ax + 1, then we can integrate the LHS of eqn 3, since each term appearing in the LHS of eqn 3 is a linear combination of the four rational functions shown above. The first of these four two can be done by substitutions s = x^2 + ax + 1 and t = x^2 - ax + 1. The fourth is similar to the third and is left as an exercise. For the third: Since a=\sqrt{2}, we get (a/2)^2 = 1/2. Completing the square gives x^2 + ax + 1 = (x + (a/2))^2 + (1/2). Then x^2 + ax + 1 = (1/2) [ (ax + 1)^2 + 1 ]. We then substitute u = ax + 1, du = a dx. Then dx (1/a) du 2 du \int ------------ = \int ---------------- = --- \int ------- x^2 + ax + 1 (1/2) [ u^2 + 1] a u^2 + 1. Since a=\sqrt{2}, we get 2/a = a. We therefore get dx \int ------------ = a \arctan(u) + C = a \arctan (ax + 1) + C. x^2 + ax + 1 ******************************************************************** Exercise 2.16, p. 93 (Mosquito flying between trains on collison course). To figure out the correct series, argue that, if the distance separating the trains at the beginning of one leg of the mosquito's journey is x, then at the end it is x/5 and the mosquito has traveled a distance of (3/5)x on that leg. On the first leg, we have x=400, so the second leg has x=400(1/5) and the third has x=400(1/5)^2, etc. The distance traveled by the mosquito is then (3/5)(400) + (3/5)(400)(1/5) + (3/5)(400)(1/5)^2 + ... = (3/5)(400)[1/(1-(1/5))] = (3/5)(400)[1/(4/5)] = (3/5)(400)(5/4) = (3/4)(400) = 300. Easy solution: Trains travel 200 miles until they collide, and they are traveling at 40 mph, so it takes 200/40 = 5 hrs until collision. The mosquito then travels for 5 hrs at 60 mph, and therefor travels a total distance of 5 x 60 = 300 mi. ******************************************************************** In Exercises 1.12, p. 85 and 2.14 p. 92, I don't care very much about the actual computation of the values of the integrals and series. I care much more about whether or not they converge. If these are assigned, it would be good to drop the part that asks for a computation of value, or to make it worth very little. ******************************************************************** Hints for Exercise 6.26, p. 74: Let f_n(x) = [(n+3) / (2n)] x + [n / (n+1)][1/x]. Then a_n=f_n(a_{n-1}). Find the number p_n > 0 such that f'_n(p_n) = 0 and show, for all n > 0, for all x > 0, that f_n(x) \ge f_n(p_n). That is, f attains its minimum at p_n. Show that f_n(p_n) --> \sqrt{2}. That is, the minimum value of f_n approaches \sqrt{2}. Conclude, for all \delta > 0, that there is some k such that, for all n\ge k, we have a_n > \sqrt{2} - \delta. Letting \delta_1=0.01 and n_1:=100, show that if |x - \sqrt{2)| < \delta_1 and if n\ge n_1, then |f'_n(x)| < 1/2. Combining this with the Mean Value Theorem, show that, for every \rho between 0 and \delta_1, there is j such that if |x - \sqrt{2}| < \rho and if i\ge j, then both |f_i(x) - f_i(\sqrt{2})| < \rho/2 and |f_i(\sqrt{2}) - \sqrt{2}| < \rho/2, in which case, |f_i(x) - \sqrt{2}| < \rho. For all n > 3, find the number q_n > 0 such that f_n(q_n) = q_n. Warning: This problem doesn't have a solution when n=3. Show, for all n > 3, that: if x \ge q_n, then f_n(x) \le x. Let S := {n | a_{n-1} < q_n}. If S is finite, then show that a_n is eventually (weakly) decreasing, i.e., that a_n has a (weakly) decreasing tail. Then, as a_n is bounded below by 0, we see that some tail of a_n is convergent, so a_n is convergent. (EXERCISE: Show that a sequence with a convergent tail is itself convergent.) Show that the limit L of a_n must satisfy both L > 0 and (1/2)L + (1/L) = L. Show that L=\sqrt{2}. So, if S is finite, then we are done. We therefore assume that S is infinite. Let \delta > 0 be given. We wish to show, for some n_0, for all n \ge n_0, that |a_n - \sqrt{2}| < \delta. Recall \delta_1 from above. Let \rho := \min{\delta,\delta_1}. Choose j such that if |x - \sqrt{2}| < \rho and i \ge j, then |f_i(x) - \sqrt{2}| < \rho. Also, as above, choose k such that: for all n\ge k, we have a_n > \sqrt{2} - \delta. Show that q_n --> \sqrt{2}. Choose m such that: for all n \ge m, we have |q_n - \sqrt{2}| < \rho. Then for all n \ge m, we have \rho - \sqrt{2} < q_n < \rho + \sqrt{2}. Let r := \max{k,m,j}. Since S is infinite, choose n_0\ge r such that n_0 + 1 \in S We wish to show, for all n \ge n_0, that |a_n - \sqrt{2}| < \delta. Since \rho = \min{\delta,\delta_1}, we get \rho \le \delta. It therefore suffices to show, for all n \ge n_0, that |a_n - \sqrt{2}| < \rho. We proceed by induction. We start by proving the inequality when n=n_0: We have n_0 \ge r \ge k, so a_{n_0} > \sqrt{2} - \rho, or \sqrt{2} - \rho < a_{n_0}. We have n_0 \ge m, so q_{n_0} < \sqrt{2} + \rho. By definition of S, as n_0 + 1 \in S, we have a_{n_0} < q_{n_0}. Then a_{n_0} < q_{n_0} < \sqrt{2} + \rho. Thus we have \sqrt{2} - \rho < a_{n_0} < \sqrt{2} + \rho, or |a_{n_0} - \sqrt{2}| < \rho. Thus the inequality holds when n=n_0. We now assume that n \ge n_0 and that |a_n - \sqrt{2}| < \rho. We wish to show that |a_{n+1} - \sqrt{2}| < \rho. Recall: if |x - \sqrt{2}| < \rho and i \ge j, then |f_i(x) - \sqrt{2}| < \rho. We have n+1 > n \ge n_0 \ge j, so, using i=n+1 and x=a_n, we have |f_{n+1}(a_n) - \sqrt{2}| < \rho. So, since a_{n+1}=f_{n+1}(a_n), we are done. ******************************************************************** Motivation for harmonic series 1 + (1/2) + (1/3) + ... Pile up n identical blocks, each of which is two units wide. Pile them up in such a way that the pile doesn't tip over. Then the center of mass of the top block lies over the second block. The center of mass of the top two blocks lies over the third block. Etc. Move the pile so that its center of mass lies over the number 2 on the real axis. Now place this pile on top of a block that extends from 0 to 2 on the real axis. The resulting pile does not tip over. The new pile will be said to be *derived* from the old pile. The center of mass of the resulting pile can be determined as follows: Let x_1 , ... , x_n be the centers of mass of the original n blocks. Then (x_1 + ... + x_n)/n is the center of mass of the pile of n blocks, which is located at 2. That is, x_1 +... + x_n = 2n. The new block has center of mass at 1. To get the center of mass of the new pile, with n+1 blocks, we compute the average of 1,x_1,...,x_n, so we get (1 + x_1 + ... + x_n) / (n+1) which is equal to (1 + 2n) / (n + 1), or (2n + 2 - 1) / (n+1), or 2 - [1/(n+1)]. Now start with a pile P_1 consisting of a single block. Derive a pile P_2 of two blocks from P_1. Then derive a pile of P_3 of three blocks from P_2. Then derive a pile of P_4 of four blocks from P_3. Etc. The center of mass of P_1 is 2 - 1. We shift P_1 over by 1 unit to the right if we want to move its center of mass to 2. So P_2 is obtained by shifting P_1 by 1 to the right and piling it on a block from 0 to 2. The center of mass of P_2 is 2 - [1/2]. We shift P_2 over by 1/2 units to the right if we want to move its center of mass to 2. So P_3 is obtained by shifting P_2 by 1/2 to the right and piling it on a block from 0 to 2. The center of mass of P_3 is 2 - [1/3]. We shift P_2 over by 1/3 units to the right if we want to move its center of mass to 2. So P_4 is obtained by shifting P_3 by 1/3 to the right and piling it on a block from 0 to 2. Etc. The rightmost point of P_1 is 2. The rightmost point of P_2 is 1+2. The rightmost point of P_3 is (1/2)+1+2. The rightmost point of P_4 is (1/3)+(1/2)+1+2. Etc. Since 1+(1/2)+(1/3)+...=\infty, we see that these piles can be made to extend as far as we like. ******************************************************************** On p. 97, there are two pictures missing. They help give a proof of Lemma 3.1, p. 96. I will show these two pictures in class, when we get to Lemma 3.1, p. 96. ******************************************************************** Note that "A & B or C" is not clear because (A & B) or C is not equivalent to A & (B or C). You might work out the truth tables to see this. On the other hand, you can see by truth tables that (A & B) & C is equivalent to A & (B & C), so that "A & B & C" is clear. Similarly, you can see by truth tables that (A or B) or C is equivalent to A or (B or C), so that "A or B or C" is clear. ******************************************************************** On p. 31, in Exercise 1.8a, there's a typo: Probably "-[\pi/6,\pi/3]" was meant to be "[-\pi/6,\pi/3]". ******************************************************************** On p.16, in the hint following Exercise 3.7, there is a sentence "Look at the list 1,2,...,p-1,p,p+1,...,p!-1,p!,p!+1 of the first p!+1 members of N". I don't think that this sentence really adds anything. If you simply delete it, I don't think anything is lost. ******************************************************************** On p. 40, there are two exercises labeled 2.21. The second of these should be 2.22. ******************************************************************** In the text, the Completeness Axiom (Axiom 2.1, on p. 32) asserts: Every nonempty subset of R which is bounded above has a least upper bound. In the text, Theorem 2.18, p.39 asserts: Every nonempty subset of R which is bounded below has a greatest lower bound. In class, I carelessly turned these two backwards. That is, as I stated the Completeness Axiom, it gives a glb to any nonempty subset of R that is bounded below, and I showed that that statement implies that any nonempty subset of R that is bounded above has a lub. In your homework, you should follow the book's exposition. Sorry for the error. ******************************************************************** Exercise 1.4(d), p. 5. Some confusion arose in class (my fault, really) and so I will drop 1.4(d) from the assignment. However 1.4(abc) are still assigned. Here's a detailed explanation: Most mathematicians equate "unless" with "if not" so that, if I say "I will go to the park unless it rains", then it is the same as saying "I will go to the park if [not (it rains)]", or, in better English, "I will go to the park if it doesn't rain", or, equivalently, "If it doesn't rain, then I'll go to the park. From this point of view, I'm not telling what I'll do if it does rain. Now, of course, if I say "I will go to the park unless it rains", then I think most people would understand that I mean "If doesn't rain, then I'll go to the park and if it does rain, then I won't go to the park", which is equivalent to "I will go to the park if and only if it doesn't rain". However, from the technical mathematical point of view, these people are jumping to conclusions, and reading into my words something that, in fact, I did not say. They just *understand* that one doesn't go to a park in the rain, and so they assume that I intend to convey this in my declaration. This was the source of the confusion. To take another example, if I say "I cannot study unless my roommate quiets down", then what I mean, really, is "If my roommate doesn't quiet down, then I cannot study", and you should *not* conclude that "If my roommate does quiet down, then I can study", because there may well be other things interfering with my ability to study. So, to be clear, from a precise mathematical point of view, "P unless Q" is equivalent to "If (not Q), then P". However, due to the confusion (on my part too, sorry), I'm dropping part (d) from the assignment, and just telling you here what the correct answer is. By the way, here's an exercise in truth tables: Show that "P unless Q" is equivalent to "P or Q". That is, the words "unless" and "or" are interchangeable. ******************************************************************** Some other useful bits of information, related to Exercise 1.4, p. 5: "P implies Q" means "if P, then Q". "a necessary condition for P is Q" means "if P, then Q". "Q is a necessary condition for P" means "if P, then Q". "a sufficient condition for Q is P" means "if P, then Q". "P is a sufficient condition for Q" means "if P, then Q". "Q if P" means "if P, then Q". "Q whenever P" means "if P, then Q". So all of the statements made above are equivalent to one another. Again, recall that "P unless Q" means "if (not Q), then P" and is equivalent to "P or Q". Some more information: "P is a necessary and sufficient condition for Q" means "P if and only if Q". "A necessary and sufficient condition for P is Q" means "P if and only if Q". "P is equivalent to Q" means "P if and only if Q". "P iff Q" is an abbreviation for "P if and only if Q". ********************************************************************